S4 Chemistry Practice Paper: Moles, Titration, Energy And Rates

S4 Chemistry Practice Paper: Moles, Titration, Energy and Rates

2026-11-12

S4 Chemistry – Moles, Titration and Rates

S4 Chemistry practice paper on moles, titration, energy changes, reaction rates and molar volume; six questions with full answers.

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Paper details　S4 Chemistry practice paper with model answers on moles, titration, energy changes and rates of reaction.

Part 1: Mole and mass calculations (20 marks)

Four questions. Show all steps, units and significant figures.

#	Question	Answer and marking notes
1	(a) Calculate the number of moles in 12.0 g of carbon (relative atomic mass 12).	Answer: moles = mass / molar mass = 12.0 / 12 = 1.00 mol. (2 marks formula, 2 marks substitution, 2 marks unit and value)
2	(b) Find the number of moles in 4.4 g of carbon dioxide (molar mass 44 g/mol) and the number of molecules.	Answer: moles = 4.4 / 44 = 0.100 mol; molecules = 0.100 x 6.02 x 10^23 = 6.02 x 10^22. (3 marks moles, 2 marks molecules)
3	(c) Find the mass of 0.250 mol of sodium carbonate (molar mass 106 g/mol).	Answer: mass = moles x molar mass = 0.250 x 106 = 26.5 g. (2 marks formula, 2 marks value and unit)
4	(d) 8.0 g of sodium hydroxide (molar mass 40 g/mol) is dissolved in water to make 250 cm3 of solution. Find the concentration in mol/dm3.	Answer: moles = 8.0 / 40 = 0.200 mol; volume = 250 cm3 = 0.250 dm3; concentration = 0.200 / 0.250 = 0.800 mol/dm3. (2 marks moles, 1 mark volume, 2 marks concentration)

Part 2: Acid-base titration calculations (18 marks)

Four questions. Write the chemical equation and show all steps.

#	Question	Answer and marking notes
5	(a) How many moles of sodium hydroxide are present in 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide solution?	Answer: moles = concentration x volume = 0.100 x 0.0250 = 2.50 x 10^-3 mol. (2 marks formula, 1 mark volume, 2 marks value)
6	(b) The solution is neutralised by 0.0500 mol/dm3 sulphuric acid. Write the balanced equation and find the volume of acid required.	Answer: H2SO4 + 2NaOH -> Na2SO4 + 2H2O; the ratio is 1:2, so moles of acid = 2.50 x 10^-3 / 2 = 1.25 x 10^-3 mol; volume = 1.25 x 10^-3 / 0.0500 = 0.0250 dm3 = 25.0 cm3. (2 marks equation, 2 marks ratio, 2 marks volume)
7	(c) State a suitable indicator and describe the colour change at the end point.	Answer: Phenolphthalein is suitable because the end point pH lies between 7 and 9, inside its range; the solution changes from pink to colourless. (2 marks indicator, 2 marks colour change)
8	(d) If the burette is not rinsed with the acid before use, will the calculated concentration of the acid be too high or too low? Explain.	Answer: The calculated concentration will be too high. Residual water dilutes the acid, so a larger volume is delivered and the calculation gives a higher concentration. (1 mark judgement, 2 marks explanation)

Part 3: Combustion and energy changes (17 marks)

Four questions. The specific heat capacity of water is 4.2 J/(g K).

#	Question	Answer and marking notes
9	(a) Burning 0.100 g of a fuel raises the temperature of 50.0 g of water by 10.0 C. Find the heat absorbed by the water.	Answer: heat = mass x specific heat capacity x temperature rise = 50.0 x 4.2 x 10.0 = 2100 J. (2 marks formula, 2 marks substitution, 1 mark unit)
10	(b) Hence find the energy released per gram of the fuel in kJ/g.	Answer: energy per gram = 2100 / 0.100 = 21000 J/g = 21.0 kJ/g. (3 marks division, 2 marks unit conversion)
11	(c) Combustion is exothermic. State the sign of the enthalpy change and explain.	Answer: The enthalpy change is negative (delta H less than zero). The reactants have more chemical energy than the products, so the excess is released as heat to the surroundings. (2 marks sign, 2 marks explanation)
12	(d) If some heat is lost to the surrounding air, will the calculated energy value be too high or too low? Explain.	Answer: It will be too low. The lost heat does not warm the water, so the measured temperature rise is smaller and the calculated energy is below the true value. (1 mark judgement, 2 marks explanation)

Part 4: Rates of reaction and catalysis (15 marks)

Four questions. Explain your answers using collision theory.

#	Question	Answer and marking notes
13	(a) When magnesium ribbon reacts with dilute hydrochloric acid, bubbles form more slowly with time. Explain.	Answer: As the reaction proceeds the acid concentration falls and the surface area of magnesium shrinks, so collisions become less frequent and gas is produced more slowly. (2 marks concentration, 2 marks surface area)
14	(b) If the same mass of magnesium powder replaces the ribbon, how does the initial rate change? Explain.	Answer: The initial rate increases. The powder has a larger surface area, so more particles are in contact with the acid and collisions per unit time increase. (2 marks change, 2 marks explanation)
15	(c) A rise of 10 C roughly doubles the rate. Explain using collision theory.	Answer: A higher temperature raises the average kinetic energy, so particles move faster and collide more often; more particles also have energy equal to or above the activation energy, raising the fraction of effective collisions. (2 marks kinetic energy and frequency, 2 marks activation energy)
16	(d) Does adding a catalyst increase the final amount of product? Explain.	Answer: No. A catalyst provides an alternative path with lower activation energy, so equilibrium is reached sooner, but the total amount of reactant is unchanged and the final yield stays the same. (1 mark judgement, 2 marks explanation)

Part 5: Molar volume of gases (15 marks)

Four questions. Take the molar volume of a gas as 24.0 dm3/mol at room conditions.

#	Question	Answer and marking notes
17	(a) Find the number of moles in 2.4 dm3 of gas at room conditions.	Answer: moles = volume / molar volume = 2.4 / 24.0 = 0.100 mol. (2 marks formula, 2 marks value)
18	(b) Find the volume occupied by 0.250 mol of carbon dioxide at room conditions.	Answer: volume = moles x molar volume = 0.250 x 24.0 = 6.0 dm3. (2 marks formula, 2 marks value and unit)
19	(c) Find the volume of 8.8 g of carbon dioxide at room conditions (molar mass 44 g/mol).	Answer: moles = 8.8 / 44 = 0.200 mol; volume = 0.200 x 24.0 = 4.8 dm3. (2 marks moles, 2 marks volume)
20	(d) Explain why different gases have the same molar volume at the same temperature and pressure.	Answer: At the same temperature and pressure, equal volumes of gases contain equal numbers of particles (Avogadro law); one mole of any substance has the same number of particles, so one mole of any gas occupies the same volume. (2 marks law, 1 mark reasoning)

Part 6: Acids, bases and pH (15 marks)

Four questions. Take the ionic product of water as 1.0 x 10^-14 at room temperature.

#	Question	Answer and marking notes
21	(a) Find the hydrogen ion concentration of hydrochloric acid at pH 2; if the solution is diluted ten times, what is the new pH?	Answer: pH 2 means [H+] = 1.0 x 10^-2 mol/dm3; after ten-fold dilution [H+] falls to 1.0 x 10^-3 mol/dm3 and pH = 3. (2 marks concentration, 2 marks new pH)
22	(b) Find the hydrogen ion and hydroxide ion concentrations of sodium hydroxide solution at pH 12.	Answer: [H+] = 1.0 x 10^-12 mol/dm3; since [H+] x [OH-] = 1.0 x 10^-14, [OH-] = 1.0 x 10^-2 mol/dm3. (2 marks each)
23	(c) 25.0 cm3 of 0.100 mol/dm3 hydrochloric acid is mixed with an equal volume of 0.100 mol/dm3 sodium hydroxide. Find the pH and state one other observable change.	Answer: The moles of acid and base are equal, so they neutralise completely to water and salt, giving pH = 7; the neutralisation is exothermic, so the mixture warms up. (2 marks pH, 2 marks temperature rise)
24	(d) Compare a pH meter with universal indicator for measuring pH.	Answer: A pH meter reads to decimal places and is more precise, without reliance on judging colour; universal indicator only gives a rough estimate and colour matching introduces error. (2 marks advantage, 1 mark comparison)

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